Let $R$ be the region in the first quadrant enclosed by the $y$ -axis, the line $y=4$, and the curve $y=x^2+2$. $y$ $x$ ${y=x^2+2}$ $ R$ $ 4$ $(0,2)$ A solid is generated by rotating $R$ about the $y$ -axis. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Answer: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=x^2+2}$ Notice the slices are horizontal, because we are rotating $R$ about the $y$ -axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=x^2+2}$ $ 4$ $(0,2)$ $r$ The radius is equal to the distance between the curve $y=x^2+2$ and the $y$ -axis. To find it, we need to solve the equation for $x$ : $x=\sqrt{y-2}$ So, for any $y$ -value, $r(y)=\sqrt{y-2}}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left(\sqrt{y-2}}\right)^2 \\\\ &=\pi (y-2) \end{aligned}$ The bottom endpoint of $R$ is at $y=2$ and the top endpoint is at $y=4$. So the interval of integration is $[2,4]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_2^4 \left[\pi (y-2)\right]dy \\\\ &=\pi\int_2^4 \left(y-2\right)dy \end{aligned}$ Let's evaluate the integral. $\pi\int_2^4 \left(y-2\right)dy=2\pi$ In conclusion, the volume of the solid is $2\pi$.